# Calculus

How would I go about solving this equation on the interval [0,2pi]

cos^2 + 2 cos x + 1 = 0

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1. cos^2(x) + 2 cos(x) + 1 = [cos(x) + 1]^2

So, cos(x) = -1 ------>

x = (2n+1)pi

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2. Muliple choice questions I have 4 options

answers.
1. pi/4, 7pi/4
2. pi/2, 3pi/2
3. pi
4. 2pi

What formula do I use ?

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3. pi

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4. thanks

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